A bug walking the edges of a tetrahedron

Question

A tetrahedron has four corners, each joined by an edge to the other three. A bug starts at one corner; each step it walks along an edge to one of the other three corners, each with probability $\tfrac13$ . It stops on reaching a chosen target corner.

What is the expected number of steps to reach the target, starting from another corner?

Accepted Answer

By symmetry, every non-target corner has the same expected count e. One step reaches the target (value 0) with probability 	frac13, or one of the two other non-target corners (value e) with probability 	frac23: e = 1 + 	frac13 (0) + 	frac23 e \;\Longrightarrow\; 	frac13 e = 1 \;\Longrightarrow\; e = 3. Takeaway. With four fully connected corners, each step hits the target with probability 	frac13, so the mean hitting time is n - 1 = 3 steps.

A bug walking the edges of a tetrahedron

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